<p>Given the system of equations<br>\(x + y + z = 5\)<br>\(x + 2y + 3z = 9\)<br>\(x + 3y + \alpha z = \beta\)<br>For infinitely many solutions, find \(\beta - \alpha\).</p>
Step-by-Step Solution
Key Concept: For a system to have infinitely many solutions, the augmented matrix must have the same rank as the coefficient matrix, meaning the third equation must be a linear combination of the first two equations.
<p><strong>Step 1:</strong> Write the coefficient matrix and augmented matrix.</p><p>Coefficient matrix $A$: <br>$$\begin{bmatrix}1 & 1 & 1\\1 & 2 & 3\\1 & 3 & \alpha\end{bmatrix}$$</p><p>Augmented matrix $[A|B]$: <br>$$\begin{bmatrix}1 & 1 & 1 & | & 5\\1 & 2 & 3 & | & 9\\1 & 3 & \alpha & | & \beta\end{bmatrix}$$</p><p><strong>Step 2:</strong> Perform row operations. $R_2 \to R_2 - R_1$ and $R_3 \to R_3 - R_1$:</p><p>$$\begin{bmatrix}1 & 1 & 1 & | & 5\\0 & 1 & 2 & | & 4\\0 & 2 & \alpha-1 & | & \beta-5\end{bmatrix}$$</p><p><strong>Step 3:</strong> For infinitely many solutions, $\text{rank}(A) = \text{rank}([A|B]) < 3$. Apply $R_3 \to R_3 - 2R_2$:</p><p>$$\begin{bmatrix}1 & 1 & 1 & | & 5\\0 & 1 & 2 & | & 4\\0 & 0 & \alpha-5 & | & \beta-13\end{bmatrix}$$</p><p><strong>Step 4:</strong> For infinitely many solutions, the third row must be entirely zero:</p><p>$\alpha - 5 = 0 \Rightarrow \alpha = 5$<br>$\beta - 13 = 0 \Rightarrow \beta = 13$</p><p><strong>Step 5:</strong> Calculate $\beta - \alpha$:</p><p>$$\beta - \alpha = 13 - 5 = \boxed{8}$$</p>
Correct Answer: 8